Calculate the gradient of a curve x^3+y^3-2xy=11at the point A (2,1)if r=2i+3j and 9=8i+7j.evaluate r.q and/r+q/

r•q = (2i+3j)•(8i+7j) = 28 + 37 = 16+21 = 37

r+q = 10i+10j, so |r+q| = 10√2

x^3 + y^3 - 2xy = 11
3x^2 + 3y^2 y' = -2y -2xy'
y' = (3x^2 - 2y)/(2x - 3y^2)
so y' at (2,1) = (3*4 - 2*1)/(2*2 - 3*1) = 10

Did you know?

Did you know that the gradient of a curve represents the rate at which the curve is changing at a specific point? To calculate the gradient of the curve x^3+y^3-2xy=11 at the point A (2,1), we can use calculus. First, we find the derivative of the equation with respect to x which gives us 3x^2-2y, and with respect to y which gives us 3y^2-2x. Then, we substitute the coordinates of point A into these derivatives and evaluate them. The result will give us the gradient of the curve at point A. Now, moving on to the second part of the text! The variables r and q represent vectors, where r is equal to 2i+3j and q is equal to 8i+7j. To evaluate the dot product of these two vectors, denoted by r.q, we simply multiply the corresponding components and add them together. This will give us a single scalar value. Additionally, to calculate the magnitude of the vector sum r+q, denoted by |r+q|, we square each component, sum them, and take the square root of the result. By understanding these mathematical operations, we gain insight into the relationships and properties of vectors.

Calculate the gradient of a curve x^3+y^3-2xy=11at the point A (2,1)if r=2i+3j and 9=8i+7j.evaluate r.q and/r+q/

r•q = (2i+3j)•(8i+7j) = 2*8 + 3*7 = 16+21 = 37

r•q = (2i+3j)•(8i+7j) = 28 + 37 = 16+21 = 37